Problem: Solve for $x$ : $2x^2 - 32x + 128 = 0$
Dividing both sides by $2$ gives: $ x^2 {-16}x + {64} = 0 $ The coefficient on the $x$ term is $-16$ and the constant term is $64$ , so we need to find two numbers that add up to $-16$ and multiply to $64$ The number $-8$ used twice satisfies both conditions: $ {-8} + {-8} = {-16} $ $ {-8} \times {-8} = {64} $ So $(x - {8})^2 = 0$ $x - 8 = 0$ Thus, $x = 8$ is the solution.